# An object with a mass of 6 kg, temperature of 240 ^oC, and a specific heat of 5 J/(kg*K) is dropped into a container with 15L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Jul 18, 2017

The water does not evaporate and the change in temperature is $= {0.11}^{\circ}$

#### Explanation:

The heat is transferred from the hot object to the cold water.

Let $T =$ final temperature of the object and the water

For the cold water, $\Delta {T}_{w} = T - 0 = T$

For the object $\Delta {T}_{o} = 240 - T$

${m}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = {m}_{w} {C}_{w} \left(\Delta {T}_{w}\right)$

${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$

${C}_{o} = 0.005 k J k {g}^{-} 1 {K}^{-} 1$

$6 \cdot 0.005 \cdot \left(240 - T\right) = 15 \cdot 4.186 \cdot T$

$240 - T = \frac{15 \cdot 4.186}{6 \cdot 0.005} \cdot T$

$240 - T = 2093 T$

$2094 T = 240$

$T = \frac{240}{2094} = {0.11}^{\circ} C$

As $T < {100}^{\circ} C$, the water does not evaporate